Question

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1. At 1000 K, the Ky value is 19.9 for the reaction: Fe2O3(s) + 3 CO (8) - 2 Fe (s) + 3 CO2(g). What are the equilibrium pressures of CO and CO2 if 10 grams of iron (III) oxide and a 0.978 atm initial partial pressure of CO are placed in a flask? a. Write the equilibrium expression, Kp, for the balanced chemical reaction: b. Determine the initial concentrations/pressure of each species present: Fe2O3(s) + 3 CO(g) = 2 Fe (s) + 3 CO2(g) Initial concentration - given, calculated or assumed to be OM The equilibrium constant expression is valid only for systems at equilibrium. When a system is not at equilibrium, the reaction quotient, Q, can be written and calculated just as K is written and calculated. The concentrations used to calculate Q are any concentration other than the equilibrium concentration. The value of Q with respect to K can indicate the direction the reaction must proceed to reach equilibrium • Q = K, the reaction is at equilibrium; • Q > K, the reaction will proceed in the direction that removes the product, moves to the left or the reverse direction; • Q< K, the reaction will proceed to the right or in the forward direction, to make product. c. Using the initial pressure to determine the value of Q and the direction of the reaction: d. Which direction will the reaction have to move to achieve equilibrium?!

Answer 1

Write the reacion as follows: Fe,(s) +3C0(8) -→2Fe(s)+300,(8) equilibrium constant, K, = 19.9 Initi al pressure of CO= 0.978 atm Equilibrium constant is defined as the ratio of product of concentrations of products raised to the powers of their stoichiometric coefficients to the product of concentrations of reactants raised to the powers of their stoichiometric coeffi cients. P Fe,0, and Fe are in pure solid state and thus their pressures are not included. | Therefore, the required solution is |K, = (Pro) (PCOM --------------- -- ------------------------- ------ (b) Write the initial pressures of the components as follows: Fe,0,(s) +3C0(8) ► 2Fe(s)+300,(8) 1(atm): - 0.978 - - - - - - - - - - - - - - - - - - (C) Calculate the value of Q as follows: O= TPY (0.978 atm = 0 ------------- ------------------------------- -- - Since, Q<K, the reaction shift to product side (forward direction). - ------ ------ -------- ------ ------ ----- ------ (e) Draw the ICE table as follows: Fe,0, (s) +300(g) →→2Fe(s)+300, (g) 1 (atm) - 0.978 C(atm): -3x + 3x E(atm): 0.978-3x + - - - - - - - - - - - - - - Substitute the corresponding valuesin equation (1) and solve for x 2,- (Pco. sy To 19.9= (0,02

(8) 3x =19.9 0.978-3x 3x = 2.71(0.978-3x) 3x = 2.65-8.13x 11.13x = 2.65 2.65 *= 11.13 x= 0.238 atm - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - Equilibrium pressure of CO= 0.978-3x = 0.978-3(0.238) = 0.978-0.714 =0.264 atm Equilibrium pressure of CO, = +3x = +3(0.238) = 0.714 atm Therefore, the equilibrium pressure are Pro = 0.264 atm and PCO, = 0.714 atm ------ ------ ----------------- Verify the values as follows: k, (Peo Substitute the equilibrium pressure values in the above equation x - (0.714 atm (0.264 atm) 0.364 Ky = 0.0184 K, = 19.8 19.9 Therefore, the calculated values are correct.